Problem 47 - Advanced Engineering Mathematics

Question

Suppose \(P(x)\) is continuous on some interval \(I\) and \(a\) is a number in \(I\). What can be said about the solution of the initial-value problem \(y^{\prime}+P(x) y=0, y(a)=0 ?\)

Step-by-Step Solution

Answer

The only solution is \( y(x) = 0 \) for all \( x \) in the interval \( I \).

Step 1: Understanding the Differential Equation

The given initial-value problem is a first-order linear homogeneous differential equation of the form \( y^{\prime} + P(x) y = 0 \). The term \( P(x) \) is a continuous function on the interval \( I \), and \( y(a) = 0 \) is the initial condition.

Step 2: Looking for a Solution

We know that a general solution to this type of differential equation can be written in the form \( y(x) = C e^{-\int P(x) \, dx} \), where \( C \) is an arbitrary constant.

Step 3: Applying the Initial Condition

The initial condition \( y(a) = 0 \) suggests substituting \( a \) into the general solution, which gives \( C e^{-\int P(x) \, dx \mid_{x=a}} = 0 \). Since the exponential term \( e^{-\int P(x) \, dx \mid_{x=a}} \) cannot be zero, this implies \( C = 0 \).

Step 4: Understanding the Implication

When \( C = 0 \), substituting back into the general solution, we have \( y(x) = 0 \cdot e^{-\int P(x) \, dx} = 0 \). This means the only solution that satisfies the initial condition is \( y(x) = 0 \) for all \( x \) in \( I \).

Key Concepts

Solution to Differential Equations

Solving differential equations is about finding the unknown function that satisfies the equation. In the context of the first-order linear homogeneous differential equation like the one given in the exercise, we look for solutions of the form \( y^{\prime} + P(x) y = 0 \).
For such equations, a general solution can be sought, typically expressed in terms of an exponential function involving an integral. That's why the general solution is often written as \( y(x) = C e^{-\int P(x) \ dx} \), where \( C \) is a constant.
- **General Solution**: Represents all possible solutions.- **Particular Solution**: Any specific instance of these general solutions.The process of solving such equations combines knowledge of calculus and algebra, leveraging integration to solve for the constant and form the complete function.

Initial-Value Problem

An initial-value problem (IVP) specifies not only a differential equation but also a particular value at which the solution should be determined. Here, the initial condition given is \( y(a) = 0 \). This initial condition helps pinpoint a specific solution from the set of general solutions.
In this exercise, applying the initial condition \( y(a) = 0 \) to the general solution highlights that the constant \( C \) in the equation becomes zero. The immediate outcome of this application is crucial since it dictates that \( y(x) = 0 \) across the defined interval. This is a unique scenario often resulting from this specific initial value where the solution collapses to the zero function.
IVPs are essential because they give additional information that is necessary to solve uniquely otherwise underdetermined systems.

Continuous Function

A continuous function like \( P(x) \) in the given differential equation is one that is continuous throughout its domain, meaning there are no sudden jumps or breaks. This property is critical for the integration process, which is part of finding the solution to the differential equation.
- **Why Continuous?**: The continuous nature of \( P(x) \) ensures the integral \( \int P(x) \, dx \) is well-defined and smooth.- **Role in Solution**: The smoothness of \( P(x) \) guarantees the solution remains valid and consistent over the interval \( I \).
If \( P(x) \) were not continuous, these guarantees would not hold, potentially leading to a non-existent or invalid solution within the interval.

Homogeneous Equations

Homogeneous differential equations like \( y^{\prime} + P(x) y = 0 \) are characterized by the equation being set to zero. The term homogeneous here means that every term in the equation can be divided by the unknown function without leaving a remainder.
- **Characteristic**: These equations are often simpler to solve due to the lack of additional non-homogeneous parts.- **Solution Implication**: Because the equation is homogeneous, the zero function, \( y(x)=0 \), becomes a solution when initial conditions similarly suggest zero, as shown from implementing \( y(a)=0 \).
Such equations highlight the principle that without any external forcing function, a zero solution is always feasible under given conditions. These equations provide foundational understanding in solving differential equations, as they align solutions with underlying mathematical properties like the superposition principle.